Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <vector>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
using Graph = vector< vector<int> >;
using ll = long long int;

const int INF = 1 << 29;
int dist_st;

vector<int> calc_dist(const Graph &G, int s, int t) {
    int N = G.size();
    vector<int> dist(N, INF);
    dist[s] = 0;
    
    queue<int> que;
    que.emplace(s);
    while(que.size()) {
        int cur = que.front(); que.pop();
        for(auto to : G[cur]) {
            if(dist[to] > dist[cur] + 1) {
                dist[to] = dist[cur] + 1;
                que.emplace(to);
            }
        }
    }

    dist_st = dist[t];
    vector<int> min_dist(N+1);
    for(int i=0; i<N; i++) if(dist[i] <= N) min_dist[ dist[i] ]++;
    return min_dist;
}

int main() {
    int N, M, s, t; cin >> N >> M >> s >> t; s--; t--;
    Graph G(N);
    for(int i=0; i<M; i++) {
        int u, v; cin >> u >> v; u--; v--;
        G[u].emplace_back(v);
        G[v].emplace_back(u);
    }

    vector<int> ds = calc_dist(G, s, t), dt = calc_dist(G, t, s);
    
    ll ans = 0;
    for(int i=0; i<=N; i++) {
        int target = dist_st - 2 - i;
        if(target < 0) continue;
        ans += 1LL * ds[i] * dt[target];
    }
    cout << ans << endl;
    return 0;
}